Answer:
A(t) = 300 -260e^(-t/50)
Step-by-step explanation:
The rate of change of A(t) is ...
A'(t) = 6 -6/300·A(t)
Rewriting, we have ...
A'(t) +(1/50)A(t) = 6
This has solution ...
A(t) = p + qe^-(t/50)
We need to find the values of p and q. Using the differential equation, we ahve ...
A'(t) = -q/50e^-(t/50) = 6 - (p +qe^-(t/50))/50
0 = 6 -p/50
p = 300
From the initial condition, ...
A(0) = 300 +q = 40
q = -260
So, the complete solution is ...
A(t) = 300 -260e^(-t/50)
___
The salt in the tank increases in exponentially decaying fashion from 40 grams to 300 grams with a time constant of 50 minutes.
To find the cross product, you can set this up as a proportion. Please see the attached paper for one of the proportions you could use.
You would then multiply the numbers that are diagonal from each other.
You would have 21(8) = 12x or
168=12x.
Answer: ![3x^2y\sqrt[3]{y}\\\\](https://tex.z-dn.net/?f=3x%5E2y%5Csqrt%5B3%5D%7By%7D%5C%5C%5C%5C)
Work Shown:
![\sqrt[3]{27x^{6}y^{4}}\\\\\sqrt[3]{3^3x^{3+3}y^{3+1}}\\\\\sqrt[3]{3^3x^{3}*x^{3}*y^{3}*y^{1}}\\\\\sqrt[3]{3^3x^{2*3}*y^{3}*y}\\\\\sqrt[3]{\left(3x^2y\right)^3*y}\\\\\sqrt[3]{\left(3x^2y\right)^3}*\sqrt[3]{y}\\\\3x^2y\sqrt[3]{y}\\\\](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B27x%5E%7B6%7Dy%5E%7B4%7D%7D%5C%5C%5C%5C%5Csqrt%5B3%5D%7B3%5E3x%5E%7B3%2B3%7Dy%5E%7B3%2B1%7D%7D%5C%5C%5C%5C%5Csqrt%5B3%5D%7B3%5E3x%5E%7B3%7D%2Ax%5E%7B3%7D%2Ay%5E%7B3%7D%2Ay%5E%7B1%7D%7D%5C%5C%5C%5C%5Csqrt%5B3%5D%7B3%5E3x%5E%7B2%2A3%7D%2Ay%5E%7B3%7D%2Ay%7D%5C%5C%5C%5C%5Csqrt%5B3%5D%7B%5Cleft%283x%5E2y%5Cright%29%5E3%2Ay%7D%5C%5C%5C%5C%5Csqrt%5B3%5D%7B%5Cleft%283x%5E2y%5Cright%29%5E3%7D%2A%5Csqrt%5B3%5D%7By%7D%5C%5C%5C%5C3x%5E2y%5Csqrt%5B3%5D%7By%7D%5C%5C%5C%5C)
Explanation:
As the steps above show, the goal is to factor the expression under the root in terms of pulling out cubed terms. That way when we apply the cube root to them, the exponents cancel. We cannot factor the y term completely, so we have a bit of leftovers.